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M1 – MOMENTS When you apply a force to a particle, there is only one point at which the force can act. With larger objects, applying the force at different points can have different effects. For example, if you close a door, it is easier if you push from the edge furthest from the hinges and much harder if you push at the point close to the hinges. The strength of the turning effect depends not only on how hard you push, but also on where and in which direction you push – in other words it depends on the line of action of the force. MOMENT OF A FORCE A force, F acts at a distance d from a point A. The turning effect of the force about the point has a magnitude of |F| x d The turning effect is called the moment of the force or the torque. As |F| is in newtons and d is in metres, the unit for the moment of a force is the newton metre (Nm). F d A You need to consider the direction (the sense) of a moment: F CLOCKWISE MOMENT ANTICLOCKWISE MOMENT d d F A A In these diagrams, the magnitude of the moment is the same, but they are in opposite senses. An anticlockwise moment is generally regarded as positive and a clockwise one as negative. In most situations you have two or more forces, each with its own turning effect. You can combine these to give an overall moment. Example 1 Forces of 10N, 15N and 18N act as shown in the rectangular lamina ABCD, with AB = 6m and BC = 4m. Find the total moment of the forces about A. 18N D C 6m 4m A 10N 15N B LAMINA: a thin plane object – in other words it has length and width but we can neglect its thickness M1 – MOMENTS Example 2 The diagram shows a triangular lamina ABC, with AB = AC = 4m. D is the mid-point of AC. Forces of 6N, 8N, 12N and P act as shown. The total moment about A is 10Nm. Find (a) the force P C 12N (b) the total moment about C. P 2m D 8N 2m 4m A EXERCISE 1 6N B M1 – MOMENTS Forces at an angle Suppose you have a rod AB of length a. You apply a force of magnitude F at B at an angle θ to the rod. You want to find the moment of the force about A. There are two possible methods: METHOD 1: The perpendicular distance from A to The line of action F is AC. F a θ A θ From the triangle ABC, AC = a sinθ. Therefore: Moment of F about A = F x AC = Fasinθ C B M1 – MOMENTS METHOD 2: Resolve F into two components. Fcosθ in the direction AB Fsinθ perpendicular to AB F Fsinθ a The first of these components is along a line Through A, so its moment about A is zero. The moment of F is produced entirely by the Second component. Hence, Moment of F about A = Fsinθ x a = Fasinθ A B θ Fcosθ Example 3 Find the total moment about the point A of the forces shown in the diagram. 12N 3.2m A 1.7m 23° 55° B 25N 20N Example 4 The diagram shows a rectangular lamina ABCD. A force of 20N is applied at C, as shown. Find the moment of this force about A. D 5.4m C 3.2m A B 35° M1 – MOMENTS Example 5 The force F = (5i + 2j)N acts at the point Q with position vector (4i + 5j)m. Find the moment of F about the point P with position vector (i + 3j)m. x 2N 6 5 5N Q 4 3 P 2 1 1 EXERCISE 2 2 3 4 5 6 y M1 – MOMENTS M1 – MOMENTS Equilibrium of parallel forces If forces act on a particle, the particle accelerates unless the forces are in equilibrium. They are in equilibrium of their resultant is zero. When forces act on a larger object, the forces may not be concurrent (in other words, acting through the same point). In this case, you need more information to decide whether they are in equilibrium. The simplest example of non-concurrent forces is when they are parallel. Such forces may have a resultant force, but they may also have a turning effect – a total moment – about some point. This system of forces has a resultant force Forces UP = 18N Forces DOWN = 12N Resultant force = 6N UP 10N There is a turning effect about the point A. Clockwise moments = 12 x 3 = 36Nm Anti-clockwise moments = 8 x 4 = 32Nm Turning effect = 4Nm clockwise A 8N 3m 1m 12N M1 – MOMENTS In general, an object acted upon by a system of parallel forces may have both linear acceleration and rotational acceleration. Both of these accelerations must be zero for the object to be in equilibrium. A system of parallel forces is in equilibrium if: The resultant of the forces is zero AND The total moment of the forces about any point is zero. Both conditions must be met if forces are to be in equilibrium. For this system Forces UP = 12N Forces DOWN = 12N Resultant force = 0N Turning effect about the point A. Clockwise moments = 12 x 3 = 36Nm Anti-clockwise moments = 9 x 4 = 36Nm Turning effect = 0Nm 3N A 9N 3m 1m 12N Therefore this system of forces is in equilibrium. A system may have a zero resultant, yet may not be in equilibrium. For this system Forces UP = 12N Forces DOWN = 12N Resultant force = 0N Turning effect about the point A. Clockwise moments = 12 x 3 = 36Nm Anti-clockwise moments = 10 x 4 = 40Nm Turning effect = 4Nm anti-clockwise 2N A 10N 3m 1m 12N Therefore this system of forces is NOT in equilibrium. A system of forces like the last one which only has a turning effect is called a couple. The moment about any point will be the same. A uniform beam means that its weight acts through the centre of the beam, which is its centre of gravity. If a beam is non-uniform its centre of gravity is not at its centre. You will either be told where the centre of gravity is or you will have to find it. M1 – MOMENTS Example 6 A uniform beam AB of mass 10kg and length 4m rests in a horizontal position on a single support C, 1m from A. The other end of the beam is supported by a vertical string as shown. 4m A 1m B C Find the reaction at the support and the tension in the string. Example 7 The diagram shows a light rod AB of length 3m. The point C divides AB in the ratio 2:1 and the forces of 6N, 7N and 8N act at A, B and C as shown. 6N PN x A 7N C D B 8N A fourth force, PN, is applied to a point D on the rod, where AD = xm, so that the system is then in equilibrium. Find the values of P and of x. M1 – MOMENTS Example 8 A uniform beam AB, of mass 20kg and length 3m, rests horizontally on two supports at A and at C, where AC = 2m. Find the reactions at both supports. A B C Example 9 A non-uniform beam AB, of length 4m and weight 500N rests horizontally on supports at A and B. The reaction at B is 140N more than the reaction at A. Find the position of the centre of gravity of the beam. M1 – MOMENTS Example 10 The diagram shows a uniform beam AB, of weight 360N and length 10m. It is supported horizontally at A and at C, where AC = 6m. An object of weight 120N is attached to the beam between B and C at a distance xm from C x A B C 120N (a) (b) Find the reactions at A and C in terms of x. Explain what happens as x increases. M1 – MOMENTS Tilting RB RA A heavy rod of mass M placed across two supports at A and B which are at the same horizontal level will be in equilibrium provided its centre of mass C is between A and B as shown. C B A Mg RB RA If an extra vertical force, F is applied to the rod at some point other than A or B the rod may or may not remain in equilibrium, depending on the magnitude and point of application of F. C B A Mg F As F increases, to remain in equilibrium the magnitude of the reaction at A must decrease to retain the equilibrium. At the time when the value of RA is just zero, the rod is said to be on the point of tilting about B. Example 11 A uniform rod AB has length 6m and mass 4kg. It is resting in equilibrium in a horizontal position on supports at points X and Y where AX = 2m and AY = 4.5m. A particle of mass M kg is placed at point C where AC = 5m. Given that the rod is on the point of tilting about Y, calculate the value of M. M1 – MOMENTS EXERCISE 3 M1 – MOMENTS M1 – MOMENTS M1 – MOMENTS M1 – MOMENTS Answers: Exercise 1 Exercise 2 Exercise 3